Question

The number of water molecules present in a drop of water (volume = 0.0018 ml) at room temperature is (density of $H_{2}O=1g/mL$)

• A. $6.023\times {10}^{19}$
• B. $1.084\times {10}^{18}$
• C. $4.84\times {10}^{17}$
• D. $6.023\times {10}^{23}$

Answer 1

The correct option is A $6.023\times {10}^{19}$

solution:

As we know that$Density=\frac{mass}{volume}$ $\Rightarrow mass=density\times volume$ Density of water $=1g/ml$ Volume of water $=0.0018mL$ $\therefore$ Mass of $0.0018mlwater=1\times 0.0018=0.0018gm$ Molar mass of water $=18gm/mol$ Number of moles in $0.0018gm$ of water $=\frac{Givenmass}{Molarmass}$= $\frac{0.0018}{18}=0.0001mole$ As we know that, Number of molecules of water in 1 mole $=6.022\times {10}^{23}molecules$ $\therefore$ Number of molecules of water in $0.0001$ mole $=6.022\times {10}^3\times 0.0001$ = $6.023\times {10}^{19}$ Hence, the number of water molecules present in a drop of water $volume=0.0018ml$ at room temperature are $6.023\times {10}^{19}$ hence the correct option : A